Existence of closed, surjective map defined in the Stone-Cech compactification

Question

I'm not so sure if my "solution" to the following problem is correct. Can you please let me know if is? or what is wrong? I'm new with this stuff so sometimes I get confused with it.

Anyway here it is:

Suppose $Y$ is an arbitrary compactification of $X$, if $\beta X$ is the Stone-Cech compactification, prove there is a continuous, surjective closed map $g: \beta X \rightarrow Y$ that equals the identity on X.

My work:

Since $Y$ is a compactification of $X$ then by definition it is a Hausdorff compact space which contains a dense copy of $X$, so we can view $X$ as a subspace of $Y$. Hence we can consider the inclusion map $i: X \rightarrow Y$. This is certainly continuous so by the universal property of extension there exists a cts extension:$g: \beta X \rightarrow Y$. So all it remains to show is that $g$ is surjective and closed. Now it is closed because $\beta X$ is compact , $Y$ is Hausdorff and any continuous map from a compact space to a Hausdorff space is closed.

Now to show $g$ is surjective I'm kinda stuck. Here's what I tried. By definition $X$ is dense in $Y$. But $\beta X$ contains $X$ so $\beta X$ is dense in $Y$ as well. Now the continuous image of a dense set is dense hence:

$\overline{g(\beta X)} = Y$

But $\beta X$ is compact and the continuous image of a compact set is compact and a compact subset in a Hausdorff space (Y) is closed so $\overline{g(\beta X)} = g(\beta X)$. Thus $g(\beta X) = Y$.

What definition of BX are you using? (Depending on the definition this property is part of the definition.) — Qiaochu Yuan, Jan 13 '11 at 23:23
Looks good to me. You've shown $g(\beta X)$ is both dense and closed in $Y$, so it must be all of $Y$. — Nate Eldredge, Jan 13 '11 at 23:32
This problem is the same as Exercise 4 of Section 38 of Munkres. In your proof, you are already using Theorem 38.4, whose proof has everything that you need for surjectivity. — Petra Axolotl, Sep 30 '23 at 21:28

score 2 · Answer 1 · answered Jan 13 '11 at 23:24

2

We know that $X \mapsto cX$ extends to a continuous map $\beta X \mapsto cX$. And $X \mapsto cX$ has a dense image. It follows that $g: \beta X \mapsto Y$ is surjective.

answered Jan 13 '11 at 23:24

NebulousReveal

13,935

by $X \mapsto cX$ you mean the inclusion map? or which map? and $cX$ is the compactification of $X$ ? – Jan 13 '11 at 23:30
@undergrad: $X \mapsto cX$ is the arbitrary compactification $Y$ of $X$. In other words, $Y: X \mapsto cX$ is the arbitrary compactification of $X$. – NebulousReveal Jan 13 '11 at 23:33

score 1 · Answer 2 · answered Nov 30 '21 at 04:26

Since $Y$ is a compactification of $X$, by the definition, there exists a cont. map $f:X\longrightarrow Y$ such that $\overline{f(X)}=Y$. By the universal property of stone-cech compactification, there exists a cont. extension $g: \beta X\longrightarrow Y$. Note that $g|_{X}=f(X)$.

By the facts that, $\beta X$ is compact, the cont. image of a compact is compact, and a compact subset of a Hausdorff space is closed, we get $g(\beta X)$ is closed in $Y$. Hence, $$Y=\overline{f(X)}\subseteq g(\beta X)\subseteq Y.$$

Existence of closed, surjective map defined in the Stone-Cech compactification

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